Statistics Assignment Solution

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Statistics Assignment Solution

Question 1(a)

Factors and the number of levels for each factor indicating any nesting present

Response Variable = Student Test Scores Factor A = School

Factor A Levels = {Lowton, Westleigh, Bedford}, a = 3

Factor B = Teacher (nested within School), b=2 levels per school

Factor B Levels = Lowton -> {T1, T2}, Westleigh -> {T3,T4}, Bedford -> {T5, T6}

Question 1(b)

𝑌𝑖𝑗𝑘 = 𝜇 + 𝛼𝑖 + 𝐵𝑗(𝑖) + 𝜖𝑘(𝑖𝑗)

where 𝐵𝑗(𝑖) ~ 𝑁(0, 𝜎𝐵2), 𝜖𝑘(𝑖𝑗) ~ 𝑁(0, 𝜎2)

And constraint Σ𝛼𝑖=03𝑖=1 for i = 1 to 3, j = 1,2 and k = 1 to 8

Question 1(c)

𝐸[ 𝑌𝑖𝑘𝑗 ] = 𝐸 [ 𝜇 + 𝛼𝑖 + 𝐵𝑗(𝑖) + 𝜖𝑘(𝑖𝑗) ]

= 𝐸 [ 𝜇 ] + 𝐸 [ 𝛼𝑖 ] + 𝐸 [ 𝐵𝑗(𝑖)] + 𝐸 [ 𝜖𝑘(𝑖𝑗) ]

= 𝜇 + 𝛼𝑖 + 0 + 0

= 𝜇 + 𝛼𝑖

= 𝜇𝑖

And

𝑉𝑎𝑟[ 𝑌𝑖𝑘𝑗 ] = 𝑉𝑎𝑟 [ 𝜇 + 𝛼𝑖 + 𝐵𝑗(𝑖) + 𝜖𝑘(𝑖𝑗) ]

= 𝑉𝑎𝑟 [ 𝜇 ] + 𝑉𝑎𝑟 [ 𝛼𝑖 ] + 𝑉𝑎𝑟 [ 𝐵𝑗(𝑖)] + 𝑉𝑎𝑟 [ 𝜖𝑘(𝑖𝑗) ]

= 0 + 0 + 𝜎𝐵2 + 𝜎2

= 𝜎𝐵2 + 𝜎2

, 2


Question 1(d)

> # 1.1a - Read in the 'School.csv' data set

> Sch1a <- read.csv("F:\\Lecture Courses\\STAT 293\\Additional Material\\Assignments &

Tests\\Assignments & Tests Data\\School.csv")

> str(Sch1a)

'data.frame': 48 obs. of 3 variables:

$ School : chr "Lowton" "Lowton" "Lowton" "Lowton" ...

$ Teacher: chr "Teacher1" "Teacher1" "Teacher1" "Teacher1" ...

$ Score : int 89 96 113 100 88 85 91 113 79 90 ...

> # 1.1b - Fitting the model.

> library(nlme)

> Sch1b <- lme(Score ~ School, random=~1|Teacher, data=Sch1a)

> # 1.1c - Obtaining the ANOVA table

> anova(Sch1b)

numDF denDF F-value p-value

(Intercept) 1 42 419.0479 <.0001

School 2 3 9.2664 0.052

Testing the null hypothesis H0: αi = 0 for all i = 1 to 3

Vs H1: αi ≠ 0 for some i = 1 to3

Fobs = 9.2664 and a p-value of 0.052 therefore there is weak evidence against the null

hypothesis.

Question 1(e)

> summary(Sch1b)